[leetcode] Gray Code

From : https://leetcode.com/problems/gray-code/

The gray code is a binary numeral system where two successive values differ in only one bit.

Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.

For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:

00 - 001 - 111 - 310 - 2

Note:
For a given n, a gray code sequence is not uniquely defined.

For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.

For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.

Solution 1: 

Binary Code ：1011 要转换成Gray Code

　　1011 = 1（照写第一位）, 1(第一位与第二位异或 1^0 = 1), 1(第二位异或第三位， 0^1=1), 0 (1^1 =0) = 1110

　　其实就等于 (1011 >> 1) ^ 1011 = 1110

class Solution {public:    vector<int> grayCode(int n) {        int size = 1<<n;        vector<int> grayCodes(size);        for (int i = 0; i < size; i++){            grayCodes[i] = (i>>1) ^i;        }        return grayCodes;    }};

Solution 2:

递归生成码表

这种方法基于格雷码是反射码的事实，利用递归的如下规则来构造：

1. 1位格雷码有两个码字,
2. (n+1)位格雷码中的前2n个码字等于n位格雷码的码字，按顺序书写，加前缀0,
3. (n+1)位格雷码中的后2n个码字等于n位格雷码的码字，按逆序书写，加前缀1.

2位格雷码3位格雷码4位格雷码4位自然二进制码

00

01

11

10

000

001

011

010

110

111

101

100

0000

0001

0011

0010

0110

0111

0101

0100

1100

1101

1111

1110

1010

1011

1001

1000

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

class Solution {public:    vector<int> grayCode(int n) {        vector<int> result;        result.push_back(0);        for(int i=0; i<n; i++){            int highest = 1<<i;            int len = result.size();            for(int i=len-1; i>=0; i--){                result.push_back(highest+result[i]);            }        }        return result;    }};